WAEC 2014 · Paper 2 · Q1✱

Given $f(y) = \log_2\left(\dfrac{y + 1}{y + 3}\right)$ :

(a)
find the domain of $f(y)$ ;
Show the answer
$\{y : y < -3 \text{ or } y > -1\}$
(b)
solve $f(y) = 1$ .
Show the answer
$y = -5$

Worked solution (try it first)

(a)

A logarithm is defined only for positive numbers, so we need $\frac{y + 1}{y + 3} > 0$ .
A fraction is positive when the top and bottom have the same sign: both positive when $y > -1$ , both negative when $y < -3$ .
The domain is $\{y : y < -3 \text{ or } y > -1\}$ .

(b)

$f(y) = 1$ means $\log_2\frac{y + 1}{y + 3} = 1$ , so $\frac{y + 1}{y + 3} = 2^1 = 2$ .
Then $y + 1 = 2y + 6$ , which gives $y = -5$ .
Check: $y = -5$ is in the domain ( $-5 < -3$ ), and $\frac{-4}{-2} = 2$ ✓.

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